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Consider two equivalent norms $|.|_{a}$ and $|.|_{b}$ on some vector space $V$. A general result then states that the norm induces the same topology on $V$.

Does that imply that $$|x|_{a} \leq |y|_{a} \text{ iff } |x|_{b} \leq |y|_{b} \tag{1}$$ ?

If not - under what conditions does two norms imply the same ordering in the senese of $(1)$ above?

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If there are two vectors $x,y$ such that $\lvert x\rvert_a=\lvert y\rvert_a=1$ and $\lvert x\rvert_b\ne\lvert y\rvert_b$, then evidently the equivalence does not hold. Therefore, in order for that to hold, it is necessary that $\lvert\bullet\rvert_b$ is constant on the vectors of $a$-norm $1$, which means that there must be some $\lambda>0$ such that $\lvert \bullet\rvert_b=\lambda\lvert \bullet\rvert_a$: namely $\lambda$ should be the constant value of the $b$-norm on the vectors of $a$-norm $1$.

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  • $\begingroup$ Nice. Very lucid! $\endgroup$
    – LudvigH
    Commented May 14, 2020 at 7:23

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